What is the tallest straw that you could, in principle, drink from

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The discussion centers on the maximum height from which water can be sipped through a straw, determined to be approximately 10 meters based on atmospheric pressure and gravity. The principle involves creating a vacuum by sucking, which allows atmospheric pressure to push the water up the straw. The calculations confirm that the pressure at the top of the straw is zero, while atmospheric pressure supports the water column. The formula used illustrates the relationship between pressure, gravity, and height, leading to the conclusion that the height of 10 meters is theoretically achievable. Overall, the consensus highlights the critical role of atmospheric pressure in this phenomenon.
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As a stunt, you want to sip some water through a very long, vertical straw. what is the tallest straw that you could, in principle, drink from in this way?
 
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Well, what do you think?
 
well i know the answer is 10 m if you round gravity to 10 m/s^2. i also know that the pressure on the top of the straw is 0 Pa. but besides that i have no clue, that's why I am asking it.
 
When you sip through a straw, what pushes the water up the straw?
 
the vacuum created by sucking your cheeks in.
i came up with this solution.
P1 + rough (g) (h1) = P2 + rough (g) (h2)
P1=0, h1=0
so P2 = rough (g)(h2)
1 atm= 1000(10)(h2)
1.01 EE5=1.0EE4(h2)
h2= 10 m

but I am not sure.
 
Looks like you've got it. It's atmospheric pressure that pushes the water up the straw. The maximum height of water that can be supported by the air will be found by setting the water pressure (\rho g h) equal to atmospheric pressure.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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